农产品现货网:已知x^2+4x+y^2-y+4(1/4)=0,求(x+y)^2-(x-y)^2的值
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要详细的解题过程
x^2+4x+y^2-y+4(1/4)=(x+2)^2+(y-1/2)^2=0 (配方 )
平方加平方等于0 x+2=0且y-1/2=0
x=-2,y=1/2
(x+y)^2-(x-y)^2=x^2+2xy+y^2-(x^2-2xy+y^2)=4xy=4*(-2)*1/2=-4
因为:x^2+4x+y^2-y+4(1/4)
=(x+2)^2+(y-1/2)^2
=0
所以:x+2=0且y-1/2=0
所以:x=-2,y=1/2
又因为:(x+y)^2-(x-y)^2
=x^2+2xy+y^2-(x^2-2xy+y^2)
=4xy
所以:4xy=4*(-2)*1/2=-4
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